A current of 12 A is passed through an electrolytic cell containing aqueous `NiSO_4` solution. Both Ni and `H_2` gas are formed at the cathode. The current efficiency is 60%. What is the mass of nickel deposited on the cathode per hour?

Options

A. `(7.883g)`
B. `(3.941g)`
C. `(5.91g)`
D. `(2.645)`

Explanation

To find the mass of nickel deposited on the cathode per hour when a current of 12 A is passed through an electrolytic cell containing an aqueous NiSO4 solution with a current efficiency of 60%, we can follow these steps:

### Step 1: Determine the Equivalent Weight of Nickel
The equivalent weight (Z) of nickel can be calculated using the formula:
\[ Z = \frac{\text{Molar Mass}}{n} \]
Where:
- Molar Mass of Ni = 58.7 g/mol
- n (number of electrons gained) = 2 (since Ni²⁺ gains 2 electrons to become Ni)

So,
\[ Z = \frac{58.7 \, \text{g/mol}}{2} = 29.35 \, \text{g/equiv} \]

### Step 2: Calculate the Total Charge (Q) Passed in One Hour
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
Where:
- I = current in amperes (12 A)
- t = time in seconds (1 hour = 3600 seconds)

So,
\[ Q = 12 \, \text{A} \times 3600 \, \text{s} = 43200 \, \text{C} \]

### Step 3: Calculate the Mass of Nickel Deposited Using Faraday's Law
According to Faraday's law, the mass (m) of the substance deposited can be calculated using the formula:
\[ m = \frac{Z \times Q \times \text{Efficiency}}{F} \]
Where:
- F = Faraday's constant = 96500 C/equiv
- Efficiency = 60% = 0.6

Substituting the values:
\[ m = \frac{29.35 \, \text{g/equiv} \times 43200 \, \text{C} \times 0.6}{96500 \, \text{C/equiv}} \]

### Step 4: Perform the Calculation
Calculating the mass:
\[ m = \frac{29.35 \times 43200 \times 0.6}{96500} \]
\[ m = \frac{760,320.0}{96500} \]
\[ m \approx 7.88 \, \text{g} \]

### Conclusion
The mass of nickel deposited on the cathode per hour is approximately **7.88 grams**.

---